The first claim is clear, so assume that $m \leq n$. Using results
we have
\begin{equation}
\begin{split}
m ! \binom{n}{m}
& = |\text{Inj}(\{ 1, \ldots, m \} \to \{ 1, \ldots, n \})| \\
& = \frac{n !}{(n - m) !}
\end{split}
\end{equation}
Dividing both sides by the nonzero quantity $m !$, the claim follows. $\square$
